Evaluate :
`lim_( x -> oo ) (( x + 2 ) ! + ( n + 1 )!) /(( x + 2 ) ! - (n + 1 ) !)`

Evaluate: lim_(x rarr oo)[(x-1)/(x-2)]^x

The value of lim_(x to oo) (1 + 2 + 3 … + n)/(n^(2)) is

Evaluate: lim_(x rarr oo)sin^(n)((2 pi n)/(3n+1)),n in N

lim_ (x rarr oo) ((1) / (n + m) + (1) / (n + 2m) + ... + (1) / (n + nm))

let lim_ (n rarr oo) ((x ^ (2) + 2x + 3 + sin pi x) ^ (n) -1) / ((x ^ (2) + 2x + 3 + sin pi x) ^ (n ) +1)

lim_ (n rarr oo) (1) / (1 + x ^ (n))

Evaluate: lim_(x rarr oo) [(4x^(n)+1)/(5x^(n)+5)] .

Statement 1: lim_ (x rarr oo) ((1 ^ (2)) / (x ^ (3)) + (2 ^ (2)) / (x ^ (3)) + (3 ^ (2)) / (x ^ (3)) + ...... + (x ^ (2)) / (x ^ (3))) = lim_ (x rarr oo) (1 ^ (2)) / (x ^ ( 3)) + lim_ (x rarr oo) (2 ^ (2)) / (x ^ (3)) + ...... + lim_ (x rarr a) (x ^ (2)) / (x ^ (3)) lim_ (x rarr a) (f_ (1) (x) + f_ (2) (x) + ... + f_ (n) (x)) = lim_ (x rarr a) f_ (1) (x) + lim_ (x rarr a) f (x) + ...... + lim_ (x rarr a) f_ (n) (x)

Evaluate lim_ (n rarr oo) n ^ (2) int _ (- (1) / (2)) ^ ((1) / (pi)) (2006sin x + 2007cos x) | x | dx

Find lim_ (n rarr oo) [x ^ (n) * (n + 1)] -: [nx ^ (n + 1)]